Introduction to Engineering thermodynamics 2 nd Edition, Sonntag and Borgnakke Solution manual PDF

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Introduction to Engineering thermodynamics 2nd Edition, Sonntag and Borgnakke Solution manual Chapter 8 Claus Borgnakke The picture is a false color thermal image of the space shuttle’s main engine. The sheet in the lower middle is after a normal shock across which you have changes in P, T and densi...

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Introduction to Engineering thermodynamics 2nd Edition, Sonntag and Borgnakke

Solution manual Chapter 8 Claus Borgnakke

The picture is a false color thermal image of the space shuttle’s main engine. The sheet in the lower middle is after a normal shock across which you have changes in P, T and density. Courtesy of NASA.

Borgnakke

CONTENT

SUBSECTION

PROB NO.

Concept-Study Guide problems 1-11 Inequality of Clausius 12-15 Entropy of a pure substance 16-24 Reversible processes 25-53 Entropy of a liquid or solid 54-65 Entropy of ideal gases 66-82 Polytropic processes 83-94 Entropy generation 95-117 Rates or fluxes of entropy 118-123 Problem solution repeated, but using the Pr and vr functions in Table A.7.2: 88, 95, 107, 112

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Borgnakke

Concept Problems

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8.1 When a substance has completed a cycle, v, u, h, and s are unchanged. Did anything happen? Explain. Yes. During various parts of the cycle work and heat transfer may be transferred. That happens at different P and T. The net work out equals the net heat transfer in (energy conservation) so dependent upon the sign it is a heat engine or a heat pump (refrigerator). The net effect is thus a conversion of energy from one storage location to another and it may also change nature (some Q got changed to W or the opposite)

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8.2 o

o

Water at 100 C, quality 50% in a rigid box is heated to 110 C. How do the properties (P, v, x, u and s) change? (increase, stay about the same, or decrease) A fixed mass in a rigid box give a constant v process. So P goes up (in the two-phase region P = Psat at given T) v stays constant. x goes up ( we get closer to the saturated vapor state see P-v diagram) u goes up (Q in and no work) s goes up (Q in) P

T

2

2 1

v

1

s

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8.3 o

Liquid water at 20 C, 100 kPa is compressed in a piston/cylinder without any heat transfer to a pressure of 200 kPa. How do the properties (T, v, u, and s) change? (increase, stay about the same, or decrease) Adiabatic dq = 0; Incompressible dv = 0: dq = T ds = 0; dw = P dv = 0 (T, v, u, and s) they are all constant. Only the pressure and enthalpy goes up.

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8.4 A reversible process in a piston/cylinder is shown in Fig. P8.4. Indicate the storage change u2 - u1 and transfers 1w2 and 1q2 as positive, zero, or negative

P 1

T

2

u=C 1 v

2 s

> 0 ; 1q2 = ∫ T ds > 0 u2 - u1 > 0 from general shape of the constant u curves. Further out in the ideal gas region the constant u curve become horizontal ( u = fct(T) only ). 1w2 =

∫ P dv

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8.5 A reversible process in a piston/cylinder is shown in Fig. P8.5. Indicate the storage change u2 - u1 and transfers 1w2 and 1q2 as positive, zero, or negative

T

P 1

1 2

2 v

s

>0; 1q2 = ∫ T ds = 0 u2 - u1 = 1q2 - 1w2 < 0

1w2 =

∫ P dv

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8.6

Air at 290 K, 100 kPa in a rigid box is heated to 325 K. How do the properties (P, v, u and s) change? (increase, stay about the same, or decrease) Rigid box: v = constant,

(P, u, and s) all increases. T

P

2 P 2

2 1

1 v

P1 s

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8.7 o

Air at 20 C, 100 kPa is compressed in a piston/cylinder without any heat transfer to a pressure of 200 kPa. How do the properties (T, v, u and s) change? (increase, about the same or decrease) T goes up, v goes down u goes up (work in, q = 0) s = constant

T

P

1 P 1 P2 2

2 1 v

s

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8.8

Carbon dioxide is compressed to a smaller volume in a polytropic process with n = 1.2. How do the properties (u, h, s, P, T) change (up, down or constant)? For carbon dioxide Table A.5 k = 1.4 so we have n < k and the process curve can be recognized in Figure 8.18. From this we see a smaller volume means moving to the left in the P-v diagram and thus also up. P up, T up and s down. As T is up so is h and u. T

P 2

2

1

1 T (n = 1) n = 1.2 v

s

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8.9

Process A: Air at 300 K, 100 kPa is heated to 310 K at constant pressure. Process B: Heat air at 1300 K to 1310 K at constant 100 kPa. Use the table below to compare the property changes. Property ∆ = v2 – v1 ∆ = h2 – h1 ∆ = s 2 – s1

∆A > ∆B

∆A ≈ ∆B √

∆A < ∆B √

√

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8.10 . A reversible heat pump has a flux of s entering as QL/TL. What can you say about the exit flux of s at TH? . . QL QH Reversible so: T = T = flux of s L H

We have the same flux of s in as out so there is no generation inside.

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8.11

An electric baseboard heater receives 1500 W of electrical power that heats the room air which looses the same amount out through the walls and windows. Specify exactly where entropy is generated in that process. a Electrical heating wire (electrical work turned into internal energy, leaving as heat transfer). b Heat transfer from hot wire to cooler room air, i.e. in the wire coverings c Room air to walls d Walls and windows, heat transfer over a finite ∆T e Walls to ambient T

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Inequality of Clausius

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8.12

Consider the steam power plant in Example 6.9 and assume an average T in the line between 1 and 2. Show that this cycle satisfies the inequality of Clausius. Solution: ⌠dQ ≤ 0 T ⌡ For this problem we have three heat transfer terms:

Show Clausius:

qb = 2831 kJ/kg, qloss = 21 kJ/kg,

qc = 2173.3 kJ/kg

⌠dq = qb – qloss – qc T T Tavg 1-2 Tc ⌡ b

2831 21 2173.3 = 573 − 568 − 318 = –1.93 kJ/kg K < 0 OK

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8.13

Assume the heat engine in Problem 7.25 has a high temperature of 1200 K and a low temperature of 400 K. What does the inequality of Clausius say about each of the four cases? Solution: ⌠ .  dQ 6 4 Cases a)  T = 1200 – 400 = – 0.005 kW/K < 0 ⌡

OK

⌠ .  dQ 6 0 b)  T = 1200 – 400 = 0.005 kW/K > 0 Impossible ⌡ ⌠ .  dQ 6 2 c)  T = 1200 – 400 = 0 kW/K ⌡

Possible if reversible

⌠ .  dQ 6 6 d)  T = 1200 – 400 = – 0.001 kW/K < 0 ⌡

OK

TH = 1200 K QH

HE

W cb

QL TL = 400 K

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8.14

Let the steam power plant in Problem 7.26 have 700oC in the boiler and 40oC during the heat rejection in the condenser. Does that satisfy the inequality of Clausius? Repeat the question for the cycle operated in reverse as a refrigerator. Solution: . QH = 1 MW

. QL = 0.58 MW

⌠ .  dQ 1000 580  T = 973 – 313 = –0.82 kW/K < 0 ⌡ Refrigerator ⌠ .  dQ 580 1000  T = 313 – 973 = 0.82 > 0 ⌡

OK

Cannot be possible

QH WT

WP, in . QL

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8.15

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequality of Clausius. . . b. W = 0 kW c. Carnot cycle a. W = 6 kW Solution: TL = 30 + 273 = 303 K TH = 250 + 273 = 523 K ; ⌠ .  dQ 6000 0 Case a)  T = 523 – 303 = 11.47 kW/K > 0 Impossible ⌡ ⌠ .  dQ 6000 6000 b)  T = 523 – 303 = –8.33 kW/K < 0 ⌡ . ⌠ .  dQ 6000 QL c)  T = 0 = 523 – 303 ⌡

OK

⇒

. 303 QL = 523 × 6 = 3.476 kW . . . W = QH – QL = 2.529 kW

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Borgnakke

Entropy of a pure substance

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8.16

Find the missing properties and give the phase of the substance a. H2O s = 7.70 kJ/kg K, P = 25 kPa h = ? T = ? x = ? b. H2O u = 3400 kJ/kg, P = 10 MPa T = ? x = ? s = ? Solution: a) Table B.1.2 sf < s < sg so two-phase and T = Tsat(P) = 64.97°C 7.70 - 0.893 x = (s – sf)/sfg = 6.9383 = 0.981 h = 271.9 + 0.981 × 2346.3 = 2573.8 kJ/kg b) Table B.1.2 u > ug => Sup.vapor Table B.1.3, x = undefined T ≅ 682°C , s ≅ 7.1223 kJ/kg K P

T

b a

b

P

a

T v

s

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8.17

Find the missing properties and give the phase of the ammonia, NH3. T = 65°C, P = 600 kPa s=?v=? a. T = 20°C, P = 100 kPa v=?s=? x=? b. 3 c. T = 50°C, v = 0.1185 m /kg s = ? x = ? P = ? a)

B.2.2 average between 60°C and 70°C v = (0.25981 + 0.26999)/2 = 0.26435 m3/kg s = (5.6383 + 5.7094)/2 = 5.6739 kJ/kgK

b)

B.2.1: P < Psat = 857.5 kPa => B.2.2 superheated vapor so x is undefined v = 1.4153 m3/kg, s = 6.2826 kJ/kgK

c)

B.2.1: v > vg = 0.06337 m3/kg => B.2.2 superheated vapor so x is undefined very close to 1200 kPa, s = 5.1497 kJ/kgK

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8.18

Find the entropy for the following water states and indicate each state on a T-s diagram relative to the two-phase region. a. 250oC, v = 0.02 m3/kg b. 250oC, 2000 kPa c. –2oC, 100 kPa Solution: 0.02 - 0.001251 a) Table B.1.1: x = = 0.38365 0.04887 s = sf + x sfg = 2.7927 + 0.38365 × 3.2802 = 4.05 kJ/kg K b) Table B.1.3: s = 6.5452 kJ/kg K c) Table B.1.5:

s = –1.2369 kJ/kg K

T

P a

c

b

a

v

c

b

s

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8.19

Find the missing properties and give the phase of the substance s=?x=? a. R-12 T = 0°C, P = 200 kPa v=?s=? b. R-134a T = −10°C, x = 0.45 c. NH3 T = 20°C, s = 5.50 kJ/kg K u = ? x = ? Solution: a) Table B.3.2, superheated vapor,

x = undefined, s = 0.7325 kJ/kg K

v = vf + xvfg = 0.000755 + 0.45 × 0.098454 = 0.04506 m3/kg

b) Table B.5.1

s = sf + xsfg = 0.9507 + 0.45 × 0.7812 = 1.3022 kJ/kg K c) Table B.2.1, s > sg => Sup.vapor Table B.2.2, x = undefined Between 400 and 500 kPa 5.5 – 5.5525 u = 1359.1 + (1353.6 – 1359.1)5.4244 – 5.5525 = 1356.8 kJ/kg

P

T

P

a, c b

b

T v

a, c s

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